Prove there exists $\lambda$ eigenvalue for $T$ and $v \notin W$ such that $(T-\lambda I)v \in W$ 
Given $V$ a vector space over $\mathbb{C}$, $T:V \rightarrow V, W\subsetneq V$, $W$ is $T$-Invariant
Prove there exists $\lambda$ eigenvalue for $T$ and $v \notin W$ such that $(T-\lambda I)v \in W.$

My idea is that $W$ should be made of vectors which are parts of jordan links (or cyclic space, I don't know what it's called), and because the dimension of $W$ is lower there must be one link (cyclic space) which misses the first vectors in $W$, so if we take the closet one to the first vector of $W$ in the link this vector should solve the question.
This is just my intuition but I failed at justifying why it is the case, so I'm looking for some guidance. Hints/feedback for my idea are welcome.
 A: It's important to note that this may be false if $W$ is not finite-dimensional. I will add a counter example as a reply to this answer, but for now, assume that $W$ is finite-dimensional
Since $W$ is $T$-invariant, it descends to a map $\overline{T}:V/W\to V/W$. Then $\overline{T}$ has an (nonzero) eigenvector $\overline{v}=v+W\in V/W$ for some $v\in V$ with eigenvalue $\lambda$. Then $\overline{T(v)}=\overline{T}(\overline{v})=\lambda\overline{v}=\overline{\lambda v}$ which implies $T(v)-\lambda v\in W$ and hence $(T-\lambda I)v\in W$. Furthermore, because $\overline{v}\neq 0$ in $V/W$, we know $v\notin W$
It remains to show that $\lambda$ is an eigenvalue of $T$ (and not just $\overline{T}$). If there exists $w'\in W$ such that $(T-\lambda I)w'=w$, then $(T-\lambda I)(v-w')=0$ and $v-w'\neq 0$, so $\lambda$ is an eigenvalue of $T$. If, on the other hand, no such $w'$ exists, then $(T-\lambda I)\vert_{W}:W\to W$ is not surjective. Since $W$ is finite-dimensional, this implies that $(T-I\lambda)\vert_{W}$ is also not injective, so there exists nonzero $w''\in W$ such that $(T-\lambda I)w''=0$ and therefore, $\lambda$ is an eigenvalue of $T$.
